Matematica discreta Esempi

$7 x + 5 y \cdot 1 = 13$ , $6 x + 4 y \cdot 1 = 19$

Passaggio 1

Move variables to the left and constant terms to the right.

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Passaggio 1.1

Moltiplica $5$ per $1$ .

$7 x + 5 y = 13$

$6 x + 4 y \cdot 1 = 19$

Passaggio 1.2

Moltiplica $4$ per $1$ .

$7 x + 5 y = 13$

$6 x + 4 y = 19$

$7 x + 5 y = 13$

$6 x + 4 y = 19$

Passaggio 2

Write the system as a matrix.

$[\begin{array}{c} 7 & 5 & 13 \\ 6 & 4 & 19 \end{array}]$

Passaggio 3

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 3.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Passaggio 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{5}{7} & \frac{13}{7} \\ 6 & 4 & 19 \end{array}]$

Passaggio 3.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & \frac{5}{7} & \frac{13}{7} \\ 6 & 4 & 19 \end{array}]$

Passaggio 3.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Passaggio 3.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{7} & \frac{13}{7} \\ 6 - 6 \cdot 1 & 4 - 6 (\frac{5}{7}) & 19 - 6 (\frac{13}{7}) \end{array}]$

Passaggio 3.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{7} & \frac{13}{7} \\ 0 & - \frac{2}{7} & \frac{55}{7} \end{array}]$

Passaggio 3.3

Multiply each element of $R_{2}$ by $- \frac{7}{2}$ to make the entry at $2, 2$ a $1$ .

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Passaggio 3.3.1

Multiply each element of $R_{2}$ by $- \frac{7}{2}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{7} & \frac{13}{7} \\ - \frac{7}{2} \cdot 0 & - \frac{7}{2} (- \frac{2}{7}) & - \frac{7}{2} \cdot \frac{55}{7} \end{array}]$

Passaggio 3.3.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{7} & \frac{13}{7} \\ 0 & 1 & - \frac{55}{2} \end{array}]$

Passaggio 3.4

Perform the row operation $R_{1} = R_{1} - \frac{5}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Passaggio 3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{7} \cdot 0 & \frac{5}{7} - \frac{5}{7} \cdot 1 & \frac{13}{7} - \frac{5}{7} (- \frac{55}{2}) \\ 0 & 1 & - \frac{55}{2} \end{array}]$

Passaggio 3.4.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{43}{2} \\ 0 & 1 & - \frac{55}{2} \end{array}]$

Passaggio 4

Use the result matrix to declare the final solution to the system of equations.

$x = \frac{43}{2}$

$y = - \frac{55}{2}$

Passaggio 5

The solution is the set of ordered pairs that make the system true.

$(\frac{43}{2}, - \frac{55}{2})$